$g(t) = \begin{cases} -2t(t+1) &, & t<6 \\\\ \dfrac{5t}{6} &, & 6\leq t\leq 12\\\\ 2t^2+3t &, & t >12\end{cases}$ $g(12)=$
Answer: The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $t={12}$. Finding the appropriate assignment rule Since $6\leq {12}\leq12$, we should use the second assignment rule $\dfrac{5t}{6}$. The answer $g({12})=\dfrac{5\cdot{12}}{6}=10$ In conclusion, $g(12)=10$.